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Doubt about calculating the solubility of lamivudine (3TC)

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Hi, I have a doubt about calculating the solubility of lamivudine (3TC). Suppose you have an excess of free base lamivudine (B) in solution. An equilibrium is generated between B in the aqueous phase and B in the solid phase, i.e. in solution and the base body. As we acidify the pH by progressive addition of blood, meanwhile, it will be dissociated, according to its acid / base balance, to dare BH +. The maximum concentration of B in water, for [BH +] ~ 0 (even if the 0 in chemistry nn exists) is precisely the intrinsic solubility, S0. Total solubility, as it is known, for pH> 4.5 is ST = + [BH +] = S0 + S0 * [H3O +] / Ka = S0 * (1+ [H3O +] / Ka). Solubility for pH <3 is ST = + [BH +] = (Kps 1: 1) ^ 1/2 + [(Kps 1: 1) ^ 1/2 * (Ka / [H3O +])] = (Kps 1: 1) ^ 1/2 * [(1 + Ka / [H3O +])].

My problem is that this does not set the equation for 3<pH <4.5. For pH = pKa = 4.2 we have that = [BH +] = 2 * (Kps 2: 1) ^ 1/3 = 2 * (5.3 * 10 ^ -4 M ^ 3) ^ 1/3 = 2 * 0.08 = 0.16 ... is it correct?

But for pH between 3 and 4.5, how is solubility calculated? I have hypothesized that, the total solubility, the mass balance, is the sum of all the concentrations of the forms in which lamivudine is present in solution, ie as a free base, conjugated acid and sale, then it could be + [ BH +] + [BHB +] or by applying the concept of cationic dimer could be + [BH +] + 2 [BHB +] (in this case it would multiply by 2 to take into account the degree of aggregation.

In both cases I don't find myself with the fact that the solubility between 4.5 and 3 goes down sale that has stoichiometry different from the usual 1: 1 In our case we have the presence of a partial sale 2: 1 that lowers the solubility. n = 2) also because , in this case, I would have to obscure the equilibrium constant of the dimer (K2), instead I only know the Kps 2: 1.

In this case K2 = Kps 2: 1?

Moreover if I consider BHB if this was in equilibrium with 2B + [H +] we have that (kps 2: 1) = (2s) ^ 2 * s = 4s ^ 3 if if BHB + was in equilibrium with B + BH + then (Kps2: 1) would be equal as * s = s ^ 2. however, there is also Cl- [BH +] [Cl -] = s ^ 3. Furthermore, what will B be equal to? If nn erro B = s only at pH = pka if I'm not mistaken. I'm going into the ball. KCl is used to assess the effect of the ion in fact a common presence of Clase plus solubility. I enclose the graphic image taken from the first edition of the book "the practice of pharmaceutical chemistry" (or better, from its translation in Italian). Thanks in advance and good day or all of you in the forum.
 

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epenguin

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You would need to re-write this as a a less confusing question. In particular specify a model that you seem to be treating, define your onstants and state their values if known, and then what the question is. Check your text eliminating obscurities like "Clase".

As far as I can make out this molecule has a protonated form whose pKa is 4.2, not surprisingly as that would be similar to cytosine, but also there is a dimeric form of limited solubility, of composition B.BH+Cl-. Is the reproduced text about this particular case and model?

I am a bit doubtful that the question has much practical importance since surely one could in all work dissolve the substance at physiological pH?
 
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You would need to re-write this as a a less confusing question. In particular specify a model that you seem to be treating, define your onstants and state their values if known, and then what the question is. Check your text eliminating obscurities like "Clase".

As far as I can make out this molecule has a protonated form whose pKa is 4.2, not surprisingly as that would be similar to cytosine, but also there is a dimeric form of limited solubility, of composition B.BH+Cl-. Is the reproduced text about this particular case and model?

I am a bit doubtful that the question has much practical importance since surely one could in all work dissolve the substance at physiological pH?
Hi, first of all thank you for answering me and for making you available to help me. My problem is that my molecule (lamivudine) shows abnormal behavior. In practice, as you well know, the molecules are much more soluble when they are in ionized form (BH +) in neutral form (B) which is why the minimum solubility value (equal to the intrinsic solubility S0) at pH> pKa +2 units and a maximum solubility value at pH <pka-2 units. In our case it is not only a sale 1: 1 (BH + Cl-) value of pH <pKa + 2 units but also a sale 2: 1 (BBH + Cl-) between 3 <pH <4.5 which causes a lowering of total solubility. The solubility of a base or its sale 1: 1 but I completely ignore how to choose to calculate the total solubility between 3 <pH <4.5 that is in the range of pH in which in solution besides B and BH + the salt is present of sale BBH + Cl-.This 2: 1 sale is for sale is not double because it has two base moles for each mole of acid (H +) rather than two moles of acid for a base. That is, it is not a polyprotic base with two protonable sites when they are of a sort of cationic dimer. The only datum of which we speak is the following: BH + but the eventual aggregation constant of the dimer, K2, or its acid constant is not mentioned; Kps 1: 1 = 0.0436 M ^ 2 (it is related to the sale BH + Cl-); Kps 2: 1 = 5.3 * 10 ^ -4 M ^ 3 (relative to the sale BBH + Cl-). Repeating the test with KCl shows that due to the effect of the ion, a common total solubility is lowered further. Furthermore, graphically we can see that S0 ~ 0.11. Please help me if you can. Thanks in advance.
 

epenguin

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You would need to re-write this as a a less confusing question. In particular specify a model that you seem to be treating, define your onstants and state their values if known, and then what the question is.

As far as I can make out this molecule has a protonated form whose pKa is 4.2, not surprisingly as that would be similar to cytosine, but also there is a dimeric form of limited solubility of composition B.BH+Cl-.
 

epenguin

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I will take a look when I have time, not promised for today. Meanwhile for other readers you need to correct "sale" → salt and "I ignore" should be "I don't know". "I ignore" will be understood as "I neglect" (trascuro).
 
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I will take a look when I have time, not promised for today. Meanwhile for other readers you need to correct "sale" → salt and "I ignore" should be "I don't know". "I ignore" will be understood as "I neglect" (trascuro).
Hi Epenguin, thanks again for your time. You're right, there are several errors in writing and I also believe in the construction of the sentence, unfortunately, due to the fact that I'm not English and I don't hang out very well :(. Unfortunately in Italy there are no well-made forums like yours and that's why "I turned to you despite being in English. Although not as soon as you can, look at it if you can. I trust you! Unfortunately, I'm trying and trying again, but I just can't manage. Thanks again for everything."
 

epenguin

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I have not made much progress on this. What the authors are trying to say is not clear to me, Probably because the quotation is incomplete – for example they refer to an equation 5 and I don't know what that is. It looks like there is a model and a mathematical development which I would need to see, so I would need both the section preceding and following your page in the original book.

I do not see any model by which it would be true that "the solubility increase with diminution of pH with slope 1".

You can probe the system with various little calculations.

For example at pH = pKa = 4.2, [ B] = [ BH+]. Then when [Cl-] = 1, Ksp2 = [ B][ BH+][Cl-] = [ B][ BH+] = [ B]2 = 5.3×10-4M2: [ B] = 0.023 M, and total B in solution is twice that so is 0.046 M which seems to be OK with the graph.

For the pH at what looks like about 2.8 it is a sufficiently good approximation that all of the B is in form BH+. Then at 1 M KCl we have Ksp1 = [ BH+][Cl-] = [ BH+] = 0.0436 M which also looks in agreement with the graph.

Have you tried a calculation like this for other points? For the points without KCl you have to take into account that, I think to good approximation in all cases, simply [ BH+] = [Cl-] Which gives some simplifications. Perhaps you did use these, I begin to see where the square roots are coming into the calculation. I mentioned two pH's where it is simple to know what you need about [ BH+] or the ratio of [ B] and [ BH+]. For the pH's at which there are not these simplifications you will need to use the Henderson-Hasselbalch equation, about the most frequently used equation in this forum or at least this section of it, to deduce this information from the pH's. And actually I am sure that the authors would have done better to have presented the data as function of HCl added, rather than or as well as pH presumably measured.

PS Also any copies should be sharp. I read that solubility constant as 5.3 × 10-4 but it is a bit fuzzy and under magnifying glass looks like 10-1, however I think that would be ridiculous and you would not be getting any precipitate from it. Sadly, any misprints cause big problems in a text like this.
 
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epenguin

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This is separate from the calculational considerations of the last post, to not overload it.

I see your profile says you are in high school – is this up-to-date? The problem is more difficult and unusual than what would normally be considered high school work. On the other hand it is nothing but an above averagely difficult application of principles that are taught in high school, so it is a valid exercise for a project perhaps.

I think, see my reaction in my first post, most biomedical researchers even would not be bothered with such detail - The most they would be bothered about would just would be how to conveniently make a stock solution and whether they can be sure it remains in solution when they pipette some into a cell culture or whatever.

However I have often looked at boxes of my medicines and noticed that they specify the hydrochloride, or the potassium salts or whatever. And thought ah yes, manufacturing convenience, mm hmm. Just in passing, because the main thought of most scientists would be the organic structure, and its interactions with proteins, enzymes, receptors and so on.

However checking up the authors to see if I could find the original book (which I couldn't, and probably it is behind a pay wall) I see that it is not a negligible problematic for manufacturers at least for formulations, and not only them, and is a small speciality with a few principles of maybe more general interest.
https://www.americanpharmaceuticalreview.com/Featured-Articles/117788-Developing-an-Appropriate-Salt-Form-for-an-Active-Pharmaceutical-Ingredient/
 
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This is separate from the calculational considerations of the last post, to not overload it.

I see your profile says you are in high school – is this up-to-date? The problem is more difficult and unusual than what would normally be considered high school work. On the other hand it is nothing but an above averagely difficult application of principles that are taught in high school, so it is a valid exercise for a project perhaps.

I think, see my reaction in my first post, most biomedical researchers even would not be bothered with such detail - The most they would be bothered about would just would be how to conveniently make a stock solution and whether they can be sure it remains in solution when they pipette some into a cell culture or whatever.

However I have often looked at boxes of my medicines and noticed that they specify the hydrochloride, or the potassium salts or whatever. And thought ah yes, manufacturing convenience, mm hmm. Just in passing, because the main thought of most scientists would be the organic structure, and its interactions with proteins, enzymes, receptors and so on.

However checking up the authors to see if I could find the original book (which I couldn't, and probably it is behind a pay wall) I see that it is not a negligible problematic for manufacturers at least for formulations, and not only them, and is a small speciality with a few principles of maybe more general interest.
https://www.americanpharmaceuticalreview.com/Featured-Articles/117788-Developing-an-Appropriate-Salt-Form-for-an-Active-Pharmaceutical-Ingredient/
 
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No, I'm not in high school but at university. I study "CFT" (chemistry and pharmaceutical technologies). It is a degree course in "Pharmacy" and is directed towards the chemical synthesis of drugs. I am preparing the applied pharmaceutical chemistry exam and my professor is still referring to a book from 1996. There are updated editions of the same book but in English, which is why they are not found in my library: absurd! In any case, checking on google books I saw that this example of 3tc is not reported on subsequent editions but for my prof is a milestone. I too could not read well if the Kps was 10 ^ -4 or 10 ^ -1 but I asked a friend to photograph the book in the library to confirm and read 10 ^ -4. I hadn't thought of proceeding by points but yours is a great idea. KCl is used to test that due to the common ion Cl- solubility further decreases. In fact, doing the calculations, the solubility without KCl is s = (Kps1: 1) ^ 1/2 = (4.36 * 10 ^ -2) ^ 1/2 = 0.21, while in the presence of KCl we have [BH +] =s if [Cl -] = 1 + s then Kps1: 1 = s * (1 + s) = s + s ^ 2 = 0.0436. Solving s = 0.04 and, as expected 0.04 <0.21. When it comes to salt 2: 1 we have that before adding KCl the [BH +] = [ B ] = [Cl -] = s for which s = (Kps2: 1) ^ 1/3 = 0.08. After KCl Kps2: 1 = s * s * (s + 1) = s ^ 2 * (s + 1). for s << 1 then s + 1 ~ 1 and Kps2: 1 = s ^ 2 * (1) = s ^ 2 ie s = (Kps2: 1) ^ 1/2 = (5.3 * 10 ^ -4) ^ 1 /2=0.023 and 0.023 <0.08. St = 2 * 0.023 = 0.046. I don't know if the procedure is correct. My doubt, however, concerns more than anything the calculation of solubility without KCl, that is in the "U" curve where solubility drops and then rises again. I do not understand what calculations have been made to obtain that andameto not only from a qualitative but quantitative point of view. Thanks again for everything. You're very kind.[/B]
 

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epenguin

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I have made a calculation for the case without added KCl.
The way I did it, maybe not the only one, is to forget about pH. I just calculate as a function of the added HCl, which in this case is the same thing as total [Cl-].
Remember I already said in #7 that it would have been more rational to present the data as function of this. And you will see in my figure a simple or simplifying feature that is not noticed, or mystified, in the presentation as function of pH.
So try and work out equations for that case as functions of [Cl-].

The curve below that represents molarity of maximum total B ([ B]+[BH+]) in solution is the lowest of the curves at every [Cl-].

This is a bit giving away the answer to your question of why the curve is of that peculiar form with kinks.
I might have hidden it from you like the book author does by giving only this lower curve and rubbing out the rest.
I hope this is sufficient enablement for you to do some calculations. That said, I cannot believe that the authors of what seems to be a standard textbook do not explain how to do these calculations, if not in the sections on either side of the extract then in earlier pages.
I have not yet done the diagrams as function of pH, and by the way this will need three separate formula for the different regions. However agreement with the text diagram appears to be better than ballpark. Neither are necessarily Evangelo just yet.

I wonder why they retired their example from their later editions - was something found wrong with it? On the other hand it is hard to imagine how they could have got the composition of their precipitates wrong.

I encourage you to update your profile which helps helpers to give best help. I am one of the people here who often answers questions on pH, buffers, ionic or other multiple equilibria, but this one with precipitates is the hardest I've met. I encourage you to complete the entire solution which could stand here as a model.

Horizontal [Cl-]. Vertical solubility


369977CC-4CC2-4065-9B01-6209DB6BCD49.jpeg
 
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I have made a calculation for the case without added KCl.
The way I did it, maybe not the only one, is to forget about pH. I just calculate as a function of the added HCl, which in this case is the same thing as total [Cl-].
Remember I already said in #7 that it would have been more rational to present the data as function of this. And you will see in my figure a simple or simplifying feature that is not noticed, or mystified, in the presentation as function of pH.
So try and work out equations for that case as functions of [Cl-].

The curve below that represents molarity of maximum total B ([ B]+[BH+]) in solution is the lowest of the curves at every [Cl-].

This is a bit giving away the answer to your question of why the curve is of that peculiar form with kinks.
I might have hidden it from you like the book author does by giving only this lower curve and rubbing out the rest.
I hope this is sufficient enablement for you to do some calculations. That said, I cannot believe that the authors of what seems to be a standard textbook do not explain how to do these calculations, if not in the sections on either side of the extract then in earlier pages.
I have not yet done the diagrams as function of pH, and by the way this will need three separate formula for the different regions. However agreement with the text diagram appears to be better than ballpark. Neither are necessarily Evangelo just yet.

I wonder why they retired their example from their later editions - was something found wrong with it? On the other hand it is hard to imagine how they could have got the composition of their precipitates wrong.

I encourage you to update your profile which helps helpers to give best help. I am one of the people here who often answers questions on pH, buffers, ionic or other multiple equilibria, but this one with precipitates is the hardest I've met. I encourage you to complete the entire solution which could stand here as a model.

Horizontal [Cl-]. Vertical solubility


View attachment 246074
Hello, even if it is hard to believe, even in the previous pages I have not found equations related to partial salts 2: 1. The only equations mentioned are: S = [HA] + [A -] = S0 (1 + Ka / [H +]); If instead I have a sodium salt Na + A- then S = (1+ [H +] / Ka) * (Kps) ^ 1/2 with Kps = [Na] [A-]; or S = [BH+] + [ B ] = S0 (1 + [H +] / Ka); S = (1 + ka / [H +]) * (Kps) ^ 1/2 with Kps = [BH+] [Cl-]. These are the only equations reported on solubility. On another text by the same author found on the internet there are the equations related to double salts (diprotics), but nothing related to partial salts. On an Avdeef book there are equations concerning the solubility of aggregates such as micelles or oligomers. For example, I was thinking of considering partial salt as a cationic dimer but I don't know K2. In any case, even if you knew (maybe you could try to design the dimer protonated with a software and calculate the Ka, because of how the equation is structured and the total solubility always increases. The equation is at this link https : //books.google.it/books? id = & n7t2nmrieGwC printsec = frontcover & hl = en # v = onepage & q & f = false
In my opinion, since solubility decreases there should be a subtractive term in the calculation of total solubility. Moreover, since the text says that between 3 and 4.5 the bottom body is constituted by the partial salt, it means that in solution there is already the maximum quantity that corresponds to its solubility. Since the aqueous concentration of the salt is in equilibrium with the body of the bottom, although the salt breaks down to restore + [BH +] + [Cl-] however its concentration remains constant. Therefore in the calculation of the total solubility 2 * [BBH + Cl-] should do 0.08 * 2 = 0.16. The problem is that by applying the formula (which I do not know if it is right in this case) + [BH +] + 2 [BBHCl] = S0 + S0 * [H +] / Ka +0.16 however, as the pH decreases, the ratio [ H +] / Ka grows and consequently the solubility increases. I don't know if I'm doing something wrong. The problem is that I don't know whether to consider them as salt as the text says or as a cationic dimer. as regards [Cl-], in the absence of KCl, is it equal to [H +] = 10 ^ -pH?
 
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Proceeding through regions, you advised me, I draw from it that the formula in the region of pH> 4.5 is St = S0 * (1+ [H +] / Ka); for pH <3 ST = (Kps1: 1) ^ 1/2 * (1+ ka / [H +]. For pH = pKa = 4.2 the [ B ] = [BH +] but ST is not equal to 2 * S0 (ie 2 times the intrinsic solubility) but 2 times the solubility of the partial salt (Kps 2: 1) ^ 1/3. to the theoretical one: in the region between 3 and 4.5 I do not know how to calculate solubility.[/B]
 

epenguin

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When I said to calculate some key simple points, I was finding this helpful just to orient myself what was going on. True at pH 4.2 the solubility is not 2S0 but in the figure in the publication you might say it was going that way, and just before getting there something else is kicking in. However I wouldn't insist on this, and let us now go directly to getting the solution.

What you have written might contain some valid points but it is very hard to disentangle them, and would be harder still to the average reader here who could not guess your idioms, e,g. 'bottom' reads strangely, and I guess means 'base' which is the same word, at least in writing, for the same chemical concept in English as in Italian, so please correct that. Also you still talk a lot about things involving [H+] which I repeat are not needed and not helpful for this stage of calculation.
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Andiamo per ordine! This definitely calls for using the homework template, as used by almost all posters in the homework section, in which you schematically set out the model you are assuming, i.e. the equilibriua, and relevant equations and quantities. And all relevant equations. You should then realise a simple thing. In solution you have only three things, or three and a bit. The bit is H+ - but its concentration will always be so smaller than that of other components that we can ignore it. (At most it might give rise to a small correction at pH near 2 but for now it is better to ignore it.) It also occurs to me, I don't know whether this is usual, but the essential relationship involving solubility products is not an equation but an inequality with a ≤ or ≥ , which helps avoid some confusions.

When you have done that we can proceed, or you might see how to take steps yourself.
 
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When I said to calculate some key simple points, I was finding this helpful just to orient myself what was going on. True at pH 4.2 the solubility is not 2S0 but in the figure in the publication you might say it was going that way, and just before getting there something else is kicking in. However I wouldn't insist on this, and let us now go directly to getting the solution.

What you have written might contain some valid points but it is very hard to disentangle them, and would be harder still to the average reader here who could not guess your idioms, e,g. 'bottom' reads strangely, and I guess means 'base' which is the same word, at least in writing, for the same chemical concept in English as in Italian, so please correct that. Also you still talk a lot about things involving [H+] which I repeat are not needed and not helpful for this stage of calculation.
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Andiamo per ordine! This definitely calls for using the homework template, as used by almost all posters in the homework section, in which you schematically set out the model you are assuming, i.e. the equilibriua, and relevant equations and quantities. And all relevant equations. You should then realise a simple thing. In solution you have only three things, or three and a bit. The bit is H+ - but its concentration will always be so smaller than that of other components that we can ignore it. (At most it might give rise to a small correction at pH near 2 but for now it is better to ignore it.) It also occurs to me, I don't know whether this is usual, but the essential relationship involving solubility products is not an equation but an inequality with a ≤ or ≥ , which helps avoid some confusions.

When you have done that we can proceed, or you might see how to take steps yourself.
What is the inequality I should set? As for the pH> pHmax1 (ie
for pH> 4.5) ST = S0 * (1 + [H3O +] / Ka) and for pH <pHmax 2 (ie for pH <3) ST = (Kps 1: 1) ^ 1/2 * (1 + Ka / [H3O +]. According to this we should consider how the solubility between pHmaX1 and pHmax2, that is for pHmax2 <pH <pHmax1. On the contrary, perhaps we should consider 4 regions instead of 3, that is to evaluate the pH trend for pHmax1 <pH <pHmin (pHmin = pKa = 4.2) and for pHmin <pH <pHmax2 If you draw a line parallel to the ordinate axis at the point pH = pka, you will get two graphs: I like to say that St = [ B ]s * (1+ [H3O +] / Ka) = [BH +]s * (1+ [H3O +] / Ka) Note that [ B ] is the solid phase of B and [BH +] is the solid phase of the conjugate acid, or rather, I correct myself because in place of [ B ]s = S0. I should consider [ B ] = [BH +] = 0.08. I miss something but I don't understand what, I'm those in the chart.[/B][/B][/B][/B]
 

epenguin

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I repeat, forget about [H+]. It does not appear in my graph.
By inequalities I mean solubility products - these are defined by equations, but the equations can get mistaken for something that has to be obeyed, instead only the corresponding inequalities have to be obeyed.
Anyway do something! Try to outline the problem in homework template form and then we can see what we are talking about.
 
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I repeat, forget about [H+]. It does not appear in my graph.
By inequalities I mean solubility products - these are defined by equations, but the equations can get mistaken for something that has to be obeyed, instead only the corresponding inequalities have to be obeyed.
Anyway do something! Try to outline the problem in homework template form and then we can see what we are talking about.
What do you mean by "outlining the model in the form of homework"? Do you want me to write in pen and set it up? If needed I can do it of course !! Indeed, now I am already at work.
 
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Hi, I was trying to think on the chart like you suggested. I made a small change in the drawing. In practice one could evaluate the total solubility (which we read on the ordinate axis) as a function of the pH (reported on the abscissa axis) as reported on the book or we could diagram the total solubility as a function of the concentration of [Cl-] on the axis of the highest abscissas as suggested by you. We know that at pH = pKa = 4.2 the [ B ] = [BH +] = [Cl-] = 0.08 while at pH = 1.7 the [BH +] = [Cl-] = 0.21. At pH = 3 St = 0.22. Should I calculate the [Cl-] graphically by intersection with the solubility curve of the base?[/B]
 

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epenguin

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Hi, I was trying to think on the chart like you suggested. I made a small change in the drawing. In practice one could evaluate the total solubility (which we read on the ordinate axis) as a function of the pH (reported on the abscissa axis) as reported on the book or we could diagram the total solubility as a function of the concentration of [Cl-] on the axis of the highest abscissas as suggested by you. We know that at pH = pKa = 4.2 the [ B ] = [BH +] = [Cl-] = 0.08 while at pH = 1.7 the [BH +] = [Cl-] = 0.21. At pH = 3 St = 0.22. Should I calculate the [Cl-] graphically by intersection with the solubility curve of the base?[/B]
That might have given you some insights but now I think we will just go in circles, talk at cross-purposes, and waste time (which I haven't got too much of) unless you set out the problem according to the official homework template of this site* the way other students do. Then we will see clearly what assumptions, equations etc we are using.


* just go to Homework Help Chemistry Biology etc. and go to start a new thread and the template will come up automatically.
 
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That might have given you some insights but now I think we will just go in circles, talk at cross-purposes, and waste time (which I haven't got too much of) unless you set out the problem according to the official homework template of this site* the way other students do. Then we will see clearly what assumptions, equations etc we are using.


* just go to Homework Help Chemistry Biology etc. and go to start a new thread and the template will come up automatically.
Hi, from your words it almost seems that I am not committing myself to solve this problem but I assure you that it is not so and I am sorry if you think this. If I wrote on this forum it is because I don't know how to deal with it and, as I have said many times, based on my studies (chemical study and pharmaceutical technologies which is a branch of pharmacy and not pure chemistry, I am not a chemist) I know just how to calculate the solubility of a base or its 1: 1 salt. I just wanted to know if there is an equation to calculate the solubility of a 2: 1 salt and I didn't want anyone to solve the exercise for me. I just wanted to know if there is a specific formula as it exists for the diprotic bases or for oligomers or micelles. I wanted to know if salt 2: 1 had to be considered as a cationic aggregate or not. As you can see it's not like I'm not just working on it, it's not as simple for me as it might seem because I'm not an expert in this field but I'm a student who is still studying these things. I don't want to learn, but if I don't understand something I can't help it (we could all understand things on the fly). If I didn't care I wouldn't post the question on a forum in English not even speaking the language well (I studied it at school level in high school). I repeat, I'm sorry because even if you think you've wasted time with me for me, the time you gave me was precious and gave me some ideas. So in spite of everything I thank you for trying to make me think. If I haven't arrived then it's my limit but I'll keep thinking about it and maybe I'll be able to find the solution. Good evening :)
 
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However rereading I noticed that I was wrong in writing but I can't change the previous message ... By mistake I wrote "I don't want to learn" instead it's exactly the opposite. I want to learn and although everyone would like to be able to understand things on the fly, most of the time you have to hit your head repeatedly to understand the concepts. I apologize again. One last thing Epenguin, always if you can tell me, I wanted to ask you a question concerning your chart. Does the green curve represent total solubility as a function of pH or as a function of the added [Cl-]? You say that the [Cl -] = HCl added but the text does not specify how much we add each time. Since HCl is a strong acid [Cl -] = [H +] = 10 ^ -pH? What do the orange and blue lines represent? In principle the slopes should be. Should I consider the curve a mathematical function? Excuse me for these questions, I don't want you to misunderstand and I don't want to know how I have to move but if I don't clarify these doubts I can't elaborate your graph well and I think that is the key to start from. Thanks in advance.
 

epenguin

Homework Helper
Gold Member
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However rereading I noticed that I was wrong in writing but I can't change the previous message ... By mistake I wrote "I don't want to learn" instead it's exactly the opposite. I want to learn and although everyone would like to be able to understand things on the fly, most of the time you have to hit your head repeatedly to understand the concepts. I apologize again. One last thing Epenguin, always if you can tell me, I wanted to ask you a question concerning your chart. Does the green curve represent total solubility as a function of pH or as a function of the added [Cl-]? You say that the [Cl -] = HCl added but the text does not specify how much we add each time. Since HCl is a strong acid [Cl -] = [H +] = 10 ^ -pH? What do the orange and blue lines represent? In principle the slopes should be. Should I consider the curve a mathematical function? Excuse me for these questions, I don't want you to misunderstand and I don't want to know how I have to move but if I don't clarify these doubts I can't elaborate your graph well and I think that is the key to start from. Thanks in advance.
The three curves represent three different inequalities. Solubility, i.e. the total molarity of dissolved B in all forms (only two!), let us call it Btot, must be on or below all of the curves. Yes, they are mathematical functions generated from a model according to my understanding of the stated problem. If your solution was at a point on one of the lowest curves and you increased the amount of B, somehow keeping everything else constant then something would precipitate- but it would be a different substance in each of the three zones.

I now think that one of the curves was mistaken, and that it should be as in the new diagram below. Co-ordinates as before
.

I recognise that you have worked on this, I am only trying to lead you a place where your work could produce conclusions faster. For this we need to use the homework template, as thousands of students who have been helped here do all the time. I noticed now this template has changed slightly but its first section is Statement of the Problem, which we already have in detail with your #1. The second, and vital, section is 'state your equations' (understand broadly, to include inequalities and chemical formulae equilibria. known quantities etc). This is what the scientist trying to solve a problem like this would be doing anyway. In homework it can reveal misunderstandings, incompletenesses or oversights, and is like an agreed basis and terminology for proceeding further.

[H+] is extremely far from equalling [Cl-]. The vast majority of protons added with the HCl go to protonating the base.

It occurs to me that you would make faster progress possibly if you first considered a simple model in which the 2:1 salt does not exist. One would expect this to be treated in a textbook, and it looks like this is what the unknown equation 5 is about, I just don't know how much they confuse issues by talking about pH. In fact I think we need to see the relevant page/s now. (When copying pages preferably use an app like HDDocScan that cleans them up.)

However whatever model you treat I don't want to see it now without your statement of relevant equations!

2B7CC992-EDB6-40E9-A301-D28325ED2660.jpeg
 
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Hi, I have a doubt about the solubility calculation of lamivudine (3TC). Suppose we have an excess of free base lamivudine (B) in solution. An equilibrium is generated between B in the aqueous phase and B in the solid phase, ie between B in solution and B as the bottom body. As we acidify the pH by gradually adding HCl, meanwhile, B will dissociate, according to its acid / base balance, to give BH +. The maximum concentration of B in water, for [BH +] ~ 0 (even if the 0 in chemistry nn exists) is precisely the intrinsic solubility, S0. Total solubility, as is known, for pH> pHmax1 (pHmax1 = 4.5) is ST = [ B ] + [BH +] = S0 + S0 * [H3O +] / Ka = S0 * (1+ [H3O + ] / Ka). The solubility for pH <pHmax2 (pHmax2 = 3) is ST = [BH +] + [ B ] = (Kps 1: 1) ^ 1/2 + [(Kps 1: 1) ^ 1/2 * (Ka / [ H3O +])] = (Kps 1: 1) ^ 1/2 * [(1 + Ka / [H3O +])].
My problem is that I ignore which equation should be set for 3 <pH <4.5. For pH = pKa = 4.2 we have that [ B ] = [BH +] and ST = [ B ] + [BH +] = 2 * (Kps 2: 1) ^ 1/3 = 2 * (5.3 * 10 ^ -4 M ^ 3) ^ 1/3 = 2 * 0.08 = 0.16 ... Is that correct?
Baut for 4.5 <pH <4.2 and 4.2 <pH <3 how is solubility calculated? I hypothesized that it is the total solubility, for the mass balance, is the sum of all the concentrations of the forms in which lamivudine is present in solution, ie as a free base, conjugated acid and salt, then it could be [ B ] + [ BH +] + [BHB +] or applying the concept of cationic dimer could be [ B ] + [BH +] + 2 [BHB +] (in this case it would multiply by 2 to take into account the degree of aggregation). I don't know K2 of dimer. Is K2= Kps2:1?
In both cases I do not find myself with the anomalous curve.
KCl is used to evaluate the effect of the common ion which further reduces solubility. I enclose the image taken from the first edition of the book "The practice of pharmaceutical chemistry" (or better, from its translation in Italian). Thanks in advance and good day or to all of you in the forum
Nuovo documento 2019-06-20 06.36.46_1.jpg
.[/B][/B][/B][/B][/B][/B]
 
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I posted the exercise in the homework section as suvited by you. Meanwhile, I look at your new chart. Thanks for everything.
 

epenguin

Homework Helper
Gold Member
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I hope we can get the problem clearly solved. I do not stand by everything I have said till now.

I'm sorry now I said your presentation was OK.I wanted to get you to write down clearly the formulae that you are using. I meant in the style of homework help presentations. Not necessarily to post in the homework forum.

Anyway I hope you will find this forum useful in the future, but the first thing you have to learn the most important thing now is how to write a question! Yours is extremely hard to understand due to the way you expose it, especially considering that parts of it are in a language that not all understand. I/we should not have to write the question for you. Surely it is possible as first section in 5 to 10 lines to explain what the system is and then your question about it. Please attend to mistakes of English and I have pointed out, E.g. I now think by " bottom body " you must mean "precipitate" also" dare" is not hard to translate.

On the second section please see what I already said about what is required. For your equations you just wrote down two different equations for ST, with no comment – what is the reader not familiar with the system and with the equations to make of that? Then these are by no means the only equations used in solving the system. In fact those formulae are only proved from other equations. With a bit of goodwill I guess the meaning of your symbols, most of which you have not defined, will be clear to the reader but the text would certainly gain in readability if you were able to use subscripts and superscripts – I don't know if you have them on your version of the site, if not readability is vastly improved by using LaTex - the site has a manual and you only need three or four things from it.

In what should be the third section, you mention something about the 2:1 complex. Please explain what you have done.You seem to know the formulae when there is only a single simple salt. Are you able to do the pH diagram for it? For your complete system with a 2:1 complex you only need another curve on top of those.

I will try to present pH curves when I can this week.
About the defects of presentation, yours is far from the only example, see from the archives

https://www.physicsforums.com/threads/single-posters-and-their-questions.887982/page-2#post-5689564

But I look forward to you getting this right.
 
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